Summation

[KindofSlow]

y=1/(x^2 + 1) find area under curve from -1 to 1 using summation formula (there are 8 rectangles under the curve).
Got all the way to sum from 1 to 8 of 4/(i^2-8i+32) and now can't figure out how to do sum with non-factorable quadratic in the denominator.
Thanks for any guidance.

 

[JeffM]

y=1/(x^2 + 1) find area under curve from -1 to 1 using summation formula (there are 8 rectangles under the curve).
Got all the way to sum from 1 to 8 of 4/(i^2-8i+32) and now can't figure out how to do sum with non-factorable quadratic in the denominator.
Thanks for any guidance.

Sorry, but I am having difficulty understanding your work and your hang-up. Can we take this from the start?

What is the width of your ith rectangle?
What is the height of that rectangle?
So the area of the ith rectangle is?

 

[galactus]

Is this a Riemann sum?.

Which method to use?.

The right endpoint method?.

The width of each rectangle is \(\displaystyle \frac{1-(-1)}{8}=\frac{1}{4}\)

How high is the rectangle at each right endpoint?.

The far left rectangle has its right side at -3/4, then -1/2, -1/4, 0, 1/4, 1/2, 3/4, 1

Enter these values in \(\displaystyle \frac{1}{x^{2}+1}\), add them up and multiply by 1/4.

 

[KindofSlow]

Supposed to use A = Sum (from 1 to n) of f(a + ((b-a)i)/n)*(b-a)/n
f(1/(x^2+1)) when x = -1+i/4
so need 4 * sum from 1 to 8 of 1/(i^2-8i+32)

Thank you

 

[galactus]

This is a Riemann right endpoint method:

You have the same thing.

\(\displaystyle \frac{1}{4}\sum_{i=1}^{8}\frac{1}{\left(-1+\frac{i}{4}\right)^{2}+1}\)

\(\displaystyle \frac{1}{4}\left(\frac{1}{(-1+\frac{1}{4})^{2}+1}+................+\frac{1}{(-1+\frac{8}{8})^{2}+1}\right)=?????\)

Now, just sum it up using 1 through 8. You should get the same as the other method I outlined in the previous post.

 

[JeffM]

Supposed to use A = Sum (from 1 to n) of f(a + ((b-a)i)/n)*(b-a)/n
f(1/(x^2+1)) when x = -1+i/4
so need 4 * sum from 1 to 8 of 1/(i^2-8i+32)

Wow

As galactus has already pointed out (b - a) / n = [1 - (- 1)] / 8 = 2/8 = 1/4 so the 4 is wrong.
Now the whole point of a Riemann sum is to add up a whole lot of smaller areas. So let's just try calculating the area of one rectangle, say rectangle 1.

Width of each rectangle = (b - a) / n = 1/4 as already discussed.
Height of rectangle 1 = f{a + [(b - a)i/n]} = f[- 1 + (2 * 1 / 8)] = f[-1 + (1/4)] = f(-3/4) = 1 / [(-3/4)[sup:3gxqu4fq]2[/sup:3gxqu4fq] + 1] = 1 / [(9 / 16) + 1] = 1 /(25/16) = 16/25.
Area of rectangle 1 = (16/25) * (1/4) = 4/25.

Now you do that arithmetic seven more times, add the eight areas up, and you have your approximation. It takes almost no time with a spreadsheet and very little time with a hand calculator.

That is the method suggested by galactus, and it works like a charm. But it does not utilize a closed form; you actually have to do the arithmetic.

Whoops, galactus beat me to it